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\title{\heiti\zihao{2} 习题9.1}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{计算下列反常积分.}
\subsection{$\int_{0}^{+\infty} \mathrm{e}^{-2 \sqrt{x}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\int_{0}^{+\infty} e^{-2 \sqrt{x}} \mathrm{~d} x=\lim _{A \rightarrow+\infty} \int_{0}^{A} e^{-2 \sqrt{x}} \mathrm{~d} x
$$

令$t=\sqrt{x},u=2t$
$$
\int_{0}^{A} e^{-2 \sqrt{x}} \mathrm{~d} x=\int_{0}^{\sqrt{A}} 2 t e^{-2 t} \mathrm{~d} t=\frac{1}{2} \int_{0}^{2 \sqrt{A}} u e^{-u} \mathrm{~d} u=\left.\frac{-x e^{-x}-e^{-x}}{2}\right|_{0} ^{2 \sqrt{A}}
$$

从而
$$
\left.\lim _{A \rightarrow+\infty} \frac{-(x+1)}{2 e^{x}}\right|_{0} ^{2 \sqrt{A}}=\frac{1}{2}
$$

\subsection{$\int_{-\infty}^{0} x^{2} \mathrm{e}^{x} \mathrm{~d} x$}
\textbf{解}\quad
$$
\int_{-\infty}^{0} x^{2} \mathrm{e}^{x} \mathrm{~d} x=\lim _{A \rightarrow-\infty} \int_{A}^{0} x^{2} e^{x} \mathrm{~d} x
$$

而
$$
\int_{A}^{0} x^{2} e^{x} \mathrm{~d} x=\left.x^{2} e^{x}\right|_{A} ^{0}-\int_{A}^{0} 2 x e^{x} \mathrm{~d} x=\left.x^{2} e^{x}\right|_{A} ^{0}-\left.2 e^{x}(x-1)\right|_{A} ^{0}
$$

对$A$取极限得:
$$
\left.\lim _{A \rightarrow-\infty} x^{2} e^{x}\right|_{A} ^{0}-\left.2 e^{x}(x-1)\right|_{A} ^{0}=2
$$

\subsection{$\int_{0}^{+\infty} \frac{\mathrm{~d} x}{\left(x^{2}+a^{2}\right)^{2}}$}
\textbf{解}\quad
$$
\int_{0}^{+\infty} \frac{\mathrm{~d} x}{\left(x^{2}+a^{2}\right)^{2}}=\lim _{A \rightarrow+\infty} \int_{0}^{A} \frac{\mathrm{~d} x}{\left(x^{2}+a^{2}\right)^{2}}
$$

令$x=at,t=\tan u$
$$
\begin{aligned}
\int_{0}^{A} \frac{\mathrm{~d} x}{\left(x^{2}+a^{2}\right)^{2}}&=\int_{0}^\frac{A}{a} \frac{\mathrm{~d} t}{a^{3}\left(t^{2}+1\right)^{2}}=\frac{1}{a^{3}} \int_{0}^{\frac{A}{a}} \frac{\mathrm{~d} t}{\left(t^{2}+1\right)^{2}}\\&=\frac{1}{a^{3}} \int_{0}^{\arctan \frac{A}{a}} \frac{\mathrm{~d} u}{\sec ^{4} u \cos ^{2} u}=\frac{1}{2 a^{3}} \int_{0}^{\arctan \frac{A}{a}} 1+\cos 2 u \mathrm{~d} u\\&=\left.\frac{1}{2 a^{3}}\left(u+\frac{\sin 2 u}{2}\right)\right|_{0} ^{\arctan \frac{A}{a}}
\end{aligned}
$$

对$A$取极限得:
$$
\left.\lim _{A \rightarrow+\infty} \frac{1}{2 a^{3}}\left(u+\frac{\sin 2 u}{2}\right)\right|_{0} ^{\arctan \frac{A}{a}}=\frac{\pi}{4a^3}
$$


\subsection{$\int_{1}^{+\infty} \frac{1}{(2 x+3)^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\int_{1}^{+\infty} \frac{1}{(2 x+3)^{2}} \mathrm{~d} x=\lim _{A \rightarrow+\infty} \int_{1}^{A} \frac{1}{(2 x+3)^{2}} \mathrm{~d} x
$$

令$2x+3=u$
$$
\int_{1}^{A} \frac{1}{(2 x+3)^{2}} \mathrm{~d} x=\frac{1}{2} \int_{5}^{2 A+3} \frac{1}{u^{2}} \mathrm{~d} u=-\left.\frac{1}{2u}\right|_{5} ^{2 A+3}
$$

对$A$取极限得:
$$
\lim _{A \rightarrow+\infty}-\left.\frac{1}{2 u}\right|_{5} ^{2 A+3}=\frac{1}{10}
$$


\subsection{$\int_{0}^{+\infty} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x$}
\textbf{解}\quad
设 $I=\int_{0}^{+\infty} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x$,则
$$
I=\int_{0}^{1} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x+\int_{1}^{+\infty} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x
$$

令$t=\frac{1}{x},$ 则有
$$
\int_{0}^{1} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x=\int_{+\infty}^{1} \frac{t \ln t}{\left(1+t^{2}\right)^{2}} \mathrm{~d} t=-\int_{1}^{+\infty} \frac{t \ln t}{\left(1+t^{2}\right)^{2}} \mathrm{~d} t=-\int_{1}^{+\infty} \frac{x \ln x}{\left(1+x^{2}\right)^{2}} \mathrm{~d} x
$$

所以 $I=0$.


\subsection{$\int_{1}^{+\infty} \frac{\arctan x}{x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\int_{1}^{+\infty} \frac{\arctan x}{x^{2}} \mathrm{~d} x=\lim _{A \rightarrow+\infty} \int_{1}^{A } \frac{\arctan x}{x^2}\mathrm{~d} x
$$

而
$$
\begin{aligned}
\int_{0}^{A} \frac{\arctan x}{x^{2}} \mathrm{~d} x &=-\left.\frac{\arctan x}{x}\right|_{1} ^{A}+\int_{1}^{A} \frac{1}{\left(x^{2}+1\right) x} \mathrm{~d} x \\
&=-\left.\frac{\arctan x}{x}\right|_{1} ^{A}+\left.\frac{1}{2} \ln \frac{x^{2}}{x^{2}+1}\right|_{1} ^{A}
\end{aligned}
$$

对$A$取极限:
$$
\lim _{A \rightarrow+\infty}-\left.\frac{\arctan x}{x}\right|_{1} ^{A}+\left.\frac{1}{2} \ln \frac{x^{2}}{x^{2}+1}\right|_{1} ^{A}=\frac{\pi}{4}-\frac{1}{2}\ln2
$$


\subsection{$\int_{0}^{+\infty} \frac{\mathrm{~d} x}{1+x^{3}}$}
\textbf{解}$1^{\circ}$\quad
$$
\int \frac{1}{1+x^{3}} \mathrm{~d} x=\int \frac{1}{3(x+1)} \mathrm{~d} x+\int \frac{-x-2}{3\left(x^{2}-x+1\right)} \mathrm{~d} x
$$

从而有
$$
\int_{0}^{+\infty} \frac{\mathrm{~d} x}{1+x^{3}}=\left.\frac{\sqrt{3}}{3} \arctan \frac{2 x-1}{\sqrt{3}}\right|_{0} ^{+\infty}=\frac{2 \sqrt{3}}{9} \pi
$$

\textbf{解}$2^{\circ}$\quad
设 $I=\int_{0}^{+\infty} \frac{\mathrm{d} x}{1+x^{3}},$ 令 $y=\frac{1}{x},$ 则
$$
I=\int_{0}^{+\infty} \frac{1}{1+x^{3}} \mathrm{~d} x=\int_{0}^{+\infty} \frac{y}{1+y^{3}} \mathrm{~d} y=\int_{0}^{+\infty} \frac{x}{1+x^{3}} \mathrm{~d} x
$$

所以
$$
\begin{aligned}
2 I &=\int_{0}^{+\infty} \frac{1+x}{1+x^{3}} \mathrm{~d} x=\int_{0}^{+\infty} \frac{1}{1-x+x^{2}} \mathrm{~d} x=\int_{0}^{+\infty} \frac{1}{\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}} \mathrm{~d} x \\
&=\left.\frac{1}{\sqrt{\frac{3}{4}}} \arctan \frac{x-\frac{1}{2}}{\sqrt{\frac{3}{4}}}\right|_{0} ^{+\infty}=\frac{2}{\sqrt{3}}\left[\frac{\pi}{2}-\arctan \left(-\frac{1}{\sqrt{3}}\right)\right] \\
&=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}+\frac{\pi}{6}\right)=\frac{4 \sqrt{3}}{9} \pi
\end{aligned}
$$

于是
$$
I=\frac{2 \sqrt{3}}{9} \pi
$$

此方法叫倒代换法,用于分母次数较高的情况.需要注意,换元的幂次不一定是$1$,可以根据题目自定.

\subsection{$\int_{0}^{+\infty} x^{3} \mathrm{e}^{-x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$
\int_{0}^{+\infty} x^{3} e^{-x^{2}} \mathrm{~d} x=\frac{1}{2} \int_{0}^{+\infty} x^{2} e^{-x^{2}} \mathrm{~d} x^{2}=\frac{1}{2} \int_{0}^{+\infty} u e^{-u} \mathrm{~d} u
$$
利用1.1结果可得积分结果为$\frac{1}{2}$

\subsection{$\int_{1}^{+\infty} \frac{1}{x^{2}+4 x+3} \mathrm{~d} x$}
\textbf{解}\quad
$$
\int_{1}^{+\infty} \frac{\mathrm{d} x}{(x+3)(x+1)}=\frac{1}{2} \int_{1}^{+\infty}\left(\frac{1}{x+1}-\frac{1}{x+3}\right) \mathrm{d} x=\lim _{A \rightarrow+\infty} \frac{1}{2} \int_{1}^{A}\left(\frac{1}{x+1}-\frac{1}{x+3}\right) \mathrm{~d} x=\frac{1}{2}\ln2
$$


\subsection{ $\int_{0}^{+\infty} \mathrm{e}^{-a x} \sin b x \mathrm{~d} x(a>0, b \neq 0)$}
\textbf{解}\quad
设 $I=\int_{0}^{+\infty} \mathrm{e}^{-a x} \sin b x \mathrm{~d} x(a>0, b \neq 0),$ 则
$$
\begin{aligned}
I &=-\frac{1}{b} \int_{0}^{+\infty} \mathrm{e}^{-a x} \mathrm{~d}(\cos b x) \\
&=\frac{1}{b}-\frac{a}{b} \int_{0}^{+\infty} \mathrm{e}^{-a x} \cos b x \mathrm{~d} x=\frac{1}{b}-\frac{a}{b^{2}} \int_{0}^{+\infty} \mathrm{e}^{-a x} \mathrm{~d}(\sin b x) \\
&=\frac{1}{b}-\frac{a^{2}}{b^{2}} \int_{0}^{+\infty} \mathrm{e}^{-a x} \sin b x \mathrm{~d} x
\end{aligned}
$$

于是
$$
I=\frac{\frac{1}{b}}{ \left(1+\frac{a^{2}}{b^{2}}\right)}=\frac{b}{a^{2}+b^{2}}
$$

\section{讨论无穷积分 $\int_{2}^{+\infty} \frac{\mathrm{d} x}{x(\ln x)^{p}}$ 的敛散性.}
\textbf{解}\quad
$$
\int_{2}^{+\infty} \frac{\mathrm{~d} x}{x(\ln x)^{p}}=\int_{\ln 2}^{+\infty} \frac{\mathrm{~d} u}{u^{p}}
$$

显然当$p>1$时收敛,$p\le1$时发散.

\section{求 $c$ 的值使得无穷积分 $\int_{0}^{+\infty}\left(\frac{2 x}{x^{2}+1}-\frac{c}{2 x+1}\right) \mathrm{d} x$ 收敛,并求出无穷积分的值.}
\textbf{解}\quad
记$f(x)=\frac{2 x}{x^{2}+1}-\frac{c}{2 x+1}$,则
$$
\int_{0}^{+\infty} f=\lim _{A \rightarrow+\infty} \int_{0}^{A} f=\lim _{A \rightarrow+\infty}\ln \left(\frac{A^{2}+1}{(2 A+1)^{\frac{c}{2}}}\right)
$$

显然$c=4$

从而有
$$
\int_{0}^{+\infty} f=\lim _{A \rightarrow+\infty}\ln \left(\frac{A^{2}+1}{(2 A+1)^2}\right)=-\ln4
$$

\section{设 $\alpha \in \mathbb{R},$ 求 $\int_{0}^{+\infty} \frac{1}{\left(x^{2}+1\right)\left(x^{\alpha}+1\right)} \mathrm{d} x$ (提示:倒代换).}
\textbf{解}\quad
令 $x=\frac{1}{t}$,
$\begin{aligned} I &=\int_{+\infty}^{0} \frac{1}{\left(1+\frac{1}{t^{2}}\right)\left(1+\frac{1}{t^{\alpha}}\right)} \frac{-1}{t^{2}} \mathrm{~d} t=\int_{0}^{+\infty} \frac{t^{\alpha}}{\left(1+t^{2}\right)\left(1+t^{\alpha}\right)} \mathrm{d} t \\ &=\int_{0}^{+\infty} \frac{x^{\alpha}}{\left(1+x^{2}\right)\left(1+x^{\alpha}\right)} \mathrm{d} x \end{aligned}$
所以 
$$
2 I=\int_{0}^{+\infty} \frac{1+x^{\alpha}}{\left(1+x^{2}\right)\left(1+x^{\alpha}\right)} \mathrm{d} x=\int_{0}^{+\infty} \frac{1}{1+x^{2}} \mathrm{~d} x=\left.\arctan x\right|_{0} ^{+\infty}=\frac{\pi}{2}
$$

故 $I=\frac{\pi}{4}$.

\section{设参数$s>0$.求$I_{n}=\int_{0}^{+\infty}e^{-sx}x^{n}\mathrm{~d} x$}
\textbf{解}\quad
令$t=sx$,则
$$
I_{n}=\frac{1}{s^{n+1}} \int_{0}^{+\infty} e^{-t} t^{n} \mathrm{~d} t=\frac{n !}{s^{n+1}}
$$

其可由$\Gamma(n)=(n-1)\Gamma(n-1)$和$\Gamma(0)=1$推得.

\section{设 $f(x)$ 是区间 $[a,+\infty)$ 上的连续函数, 且 $\int_{a}^{+\infty} f(x) \mathrm{d} x$ 收敛. 证明:存在数列 $\left\{x_{n}\right\}$ 使 得 $\lim _{n \rightarrow \infty} x_{n}=+\infty,$ 且 $\lim _{n \rightarrow \infty} f\left(x_{n}\right)=0$}
\textbf{证}\quad
由于$f(x)\in C[a,+\infty)$,所以其存在原函数$F(x)$且原函数可导.由题设有$\lim _{x \rightarrow+\infty} F(x)=A$

则由柯西收敛准则,$\exists N,s.t.\forall\varepsilon,\forall x^{\prime},x^{\prime\prime}>N,|F(x^{\prime})-F(x^{\prime\prime})|<\varepsilon$

而由柯西中值定理,$\exists\xi \in (x^{\prime},x^{\prime\prime}),F^{\prime}(\xi)=\frac{F(x^{\prime})-F(x^{\prime\prime})}{x^{\prime}-x^{\prime\prime}}$

而显然可取$x^{\prime\prime}>x^{\prime}+1$,从而$\left|\frac{F(x^{\prime})-F(x^{\prime\prime})}{x^{\prime}-x^{\prime\prime}}\right|<\varepsilon$,从而找到$-\varepsilon<f(\xi)=F^{\prime}(\xi)<\varepsilon$.(因为$f(x)$连续)

只要取$x_{n+1}=2+x_{n}$即可得到数列,其极限趋于正无穷,而在这些点的导数趋于0.




\end{document}
\subsection{}
\textbf{解}\quad

\subsection{}
\textbf{证}\quad

\textbf{\textcolor{red}{注}}\quad